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\begin{flashcard}[G-invariant-inner-product]
\begin{definition*}[$G$-invariant inner product]
  \glsnoundefn{Ginv_ip}{$G$-invariant inner
  product}{$G$-invariant inner products}
  \cloze{
  A Hermitian inner product on a \gls{rep}
  $V$ of $G$ is \cloze{$G$-invariant} if $(gx, gy)
  = (x, y)$ for all $g \in G$, $x, y \in V$.

  \fcemph{Equivalently} if $(gx, gx) = (x, x)
  ~\forall g \in G, x \in V$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[unitary-iff-G-invariant-inner-product]
\begin{proposition*}
  A \gls{rep} $(\rho, V)$ of $G$ is unitary
  if and only if \cloze{$V$ has a \Ginvip.}
\end{proposition*}

\begin{proof}
  \cloze{
  If $(\rho, V)$ is contrary let $e_1, \ldots,
  e_n$ be a basis with respect to which each
  $\rho(g) \in U(n)$. Now,
  \[ \left( \sum_i \lambda_i e_i, \sum_j \mu_j e_j
  \right) = \sum_i \ol{\lambda_i} \mu_i \]
  is a \Ginvip\ on $V$.

  Conversely, if $V$ has a \Ginvip
  $(\bullet, \bullet)$, we can find an
  orthonormal basis $v_1,
  \ldots, v_n$ of $V$ with respect to $(\bullet,
  \bullet)$. Then
  \[ \left( \sum_i \lambda_i v_i, \sum_j \mu_j v_j
  \right) = \sum_i \ol{\lambda_i} \mu_i \qquad
  \forall \lambda, \mu \in \CC^n \]
  i.e. $(\bullet, \bullet)$ is the standard inner
  product with respect to $v_1, \ldots, v_n$, so
  since it is \glsref[Ginv_ip]{$G$-invariant} each $\rho(g)$ is
  unitary with respect to this basis.
  }
\end{proof}
\end{flashcard}

Note \glspl{subrep} of unitary \glspl{rep}
are thus unitary since we can restrict a
\Ginvip.

\begin{flashcard}[unitary-rep-implies-completely-reducible]
\begin{theorem*}
  \cloze{Suppose $(\rho, V)$ is a unitary representation
  of a group $G$. Then every subrepresentation $W$
  of $V$ has a $G$-invariant complement.} Thus $V$
  is completely reducible.
\end{theorem*}

\begin{proof}
  \cloze{
  Since $V$ is unitary it has a $G$-invariant
  inner product $(\bullet, \bullet)$. If $W$ is a
  subrepresentation then
  \[ W^\perp = \{v \in V \mid (v, w) = 0 ~\forall
  w \in W\} \]
  is a vector space complement to $W$ in $V$ by
  standard linear algebra. Moreover, if $g \in G$,
  $v \in W^\perp$ and $w \in W$ then
  \[ (gv, w) = (v, g^{-1}w) = 0 \]
  since $g^{-1} w \in W$, so $gv \in W^\perp$ and
  $W^\perp$ is a subrepresentation as required.

  \fcscrap{
  The last part follows from lemma proved last
  time.}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[maschkes-thm]
\begin{theorem*}[Maschke's Theorem]
  \refstepcounter{customtheorem}
  \label{maschkes_thm}
  \cloze{Let $G$ be a finite group and $(\rho, V)$ is a
  \gls{rep} of $G$ over $k$, a field of
  characteristic zero. Suppose $W \le V$ is a
  \gls{subrep}. Then $W$ has a \Ginv
  complement in $V$. In particular, $V$ is
  \gls{comp_red}.}
\end{theorem*}
\end{flashcard}

\refsteplabel[key idea]{lec5_key_idea}
\textbf{Key idea:} If $(\rho, V)$ is a
\gls{rep} of a finite group $G$ over a field
$k$ then for all $v \in V$,
\[ \sum_{g \in G} g \cdot v \in V^G = \{v \in V
\mid g \cdot v = v ~\forall g \in G\} \le V  \]

\begin{proof}[Proof of Key idea]
  If $h \in G$,
  \[ h \left( \sum_{g \in G} g \cdot v \right) =
  \sum_{g \in G} (hg) \cdot v = \sum_{g' \in G} g'
  \cdot v \]
  since $G \to G$, $g \mapsto hg$ is a permutation
  of $G$ and $h : V \to V$ is linear.
\end{proof}

\begin{flashcard}[weyls-unitary-trick]
\begin{proposition*}[Weyl's unitary trick]
  \cloze{If $V$ is a $\CC$-\gls{rep} of a finite
  group $G$ then $V$ has a \Ginvip.
  Thus \nameref{maschkes_thm} is true
  over $\CC$.}
\end{proposition*}

\begin{proof}
  \cloze{
  Pick any Hermitian inner product $\langle
  \bullet, \bullet \rangle$ on $V$. Then we can
  define a new inner product on $V$ via
  \[ (x, y) = \sum_{g \in G} \langle gx, gy
  \rangle \]
  It is easy to see that $(\bullet, \bullet)$ is a
  Hermitian inner product because $\langle
  \bullet, \bullet \rangle$ is since, for example,
  if $a, b \in \CC$ and $x, y, z \in V$ then
  \begin{align*}
    (x, ay + bz)
    &= \sum_{g \in G} \langle gx, g(ay + bz)
    \rangle \\
    &= \sum_{g \in G} \langle gx, a g(y) + bg(z)
    \rangle \\
    &= \sum_{g \in G} (a \langle gx, gy \rangle +
    b \langle gx, gz \rangle) \\
    &= a (x, y) + b(x, z)
  \end{align*}
  But now if $h \in G$ and $x, y \in V$,
  \[ (hx, hy) = \sum_{g \in G} \langle ghx, ghy
  \rangle = \sum_{g' \in G} \langle g'x, g'y
  \rangle = (x, y) \]
  since $g \mapsto gh$ is a permutation of $G$.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  This proof can be phrased as follows:
  \begin{enumerate}[(i)]
    \item $\Herm(V) = \{\text{Hermitian
      sesquilinear forms}\}$ is naturally an
      $\RR$-vector space.
    \item $G \to \Aut(\Herm(V))$, $g \cdot
      (\bullet, \bullet)|_{(x, y)} = (g^{-1} x,
      g^{-1} y)$ defines an $\RR$-linear
      \gls{rep} of $G$.
    \item All $\RR^{>0}$-linear combinations of
      positive definite elements of $\Herm(V)$ are
      positive definite.
    \item The \nameref{lec5_key_idea} transforms an inner product
      into a \glsref[Ginv_ip]{$G$-invariant} one.
  \end{enumerate}
  It follows that studying $\CC$-\glspl{rep}
  of a finite group is the same as studying
  unitary \glspl{rep} of the group.
\end{remark*}

\begin{flashcard}[finite-subgp-of-GLnC-conj-to-subgp-of-Un]
\begin{corollary*}
  Every finite subgroup of $\GL_n(\CC)$ is
  conjugate \cloze{to a subgroup of $U(n)$.}
\end{corollary*}

\begin{proof}
  \cloze{If $G \le \GL_n(\CC)$ then the inclusion map
  $\rho : G \to \GL_n(\CC)$ is a \gls{rep}.
  By the unitary trick, there exists a basis for
  $\CC^n$ with respect to which each $\rho(g) \in
  U(n)$, i.e. $\exists P \in \GL_n(\CC)$ such that
  $\forall g \in G$, $P \rho(g) P^{-1} \in U(n)$.}
\end{proof}
\end{flashcard}

We now generalise to all $\characteristic 0$
fields.

\begin{proof}[Proof of \nameref{maschkes_thm}]
  Idea: if $\pi : V \to V$ is a projection, i.e.
  $\pi^2 = \pi$, then $V = \Im \pi \oplus \ker
  \pi$ as vector spaces. If $\pi$ is $G$-linear
  then $\ker \pi$ and $\Im \pi$ are
  \glspl{subrep}. So $\Im \pi$ has a
  \Ginv\ complement. So we pick a
  projection onto $W$ and average it.
  
  Let $\pi : V \to V$ be any $k$-linear projection
  onto $W$ (so $\pi(w) = w$ for all $w \in W$ and
  $\Im \pi = W$). Recall that $\Hom_k(V, V)$ is a
  \gls{rep} of $G$ via $g \cdot \varphi = g
  \circ \varphi \circ g^{-1}$. Let $\pi^G =
  \frac{1}{|G|} \sum_{g \in G} (g \circ \pi) \in
  \Hom_G(V, V)$ by the \nameref{lec5_key_idea}.

  Moreover, $\Im \pi^G \le W$ since $g \circ \pi
  g^{-1}(v) \in W$ for all $v \in V, g \in G$ and
  \[ \pi^G(v) = \sum_{g \in G} \frac{1}{|G|} g
  \circ \pi \circ g^{-1}(v) \]
  and if $w \in W$ then
  \[ \pi^G(w) = \frac{1}{|G|} \sum_{g \in G} g
  \circ \pi \circ g^{-1}(w) = \frac{1}{|G|}
  \sum_{g \in G} g \circ g^{-1} (w) = w \]
  since $g^{-1}(w) \in W ~\forall g \in G, w \in
  W$. So $\pi^G$ is a $G$-invariant linear
  projection onto $W$ and $\ker \pi$ is a
  \Ginv\ complement to $W$ in $V$.
\end{proof}

\begin{remark*}
  \refstepcounter{customremark}
  \label{lecture5_last_remark}
  \phantom{}
  \begin{enumerate}[(1)]
    \item We can explicitly compute $\pi^G$ and
      $\ker \pi^G$ via formula
      \[ \pi^G = \frac{1}{|G|} \sum_{g \in G} g
      \cdot \pi \]
    \item Notice we only used $\characteristic(k)
      = 0$ when we divided by $|G|$. So in fact
      the result holds whenever
      $\characteristic(k) \nmid |G|$.
    \item As an extension of our
      \nameref{lec5_key_idea}, for any
      $G$-\gls{rep} $V$ (and $\characteristic
      k \nmid |G|$),
      \refsteplabel[projection]{UG_projection}
      \[ \pi : v \mapsto \frac{1}{|G|} \sum_{g \in
      G} gv \]
      is a projection in $\Hom_G(V, V)$ onto
      $V^G$. Notice $\dim V^G = \Trace \pi =
      \frac{1}{|G|} \sum_{g \in G} \Trace(g)$ since
      $\Trace$ is linear.
  \end{enumerate}
\end{remark*}
